What is inverse oflaplace-1{(15)/(s^2+4s+13)} ?

The answer to inverse oflaplace-1{(15)/(s^2+4s+13)} is -5e^{-2t}sin(3t)

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inverse oflaplace-1{(15)/(s^2+4s+13)}

inverse laplace

- 1 {\frac{15}{s ^{2} + 4 s + 13}}

- 5 e^{- 2 t} sin (3 t)

Solution steps

L^{- 1} {- 1 \cdot {\frac{15}{s ^{2} + 4 s + 13}}}

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

= - 1 \cdot L^{- 1} {{\frac{15}{s ^{2} + 4 s + 13}}}

L^{- 1} {{\frac{15}{s ^{2} + 4 s + 13}}} : 5 e^{- 2 t} sin (3 t)

= - 1 \cdot 5 e^{- 2 t} sin (3 t)

Refine

- 15 e^{- 2 t} sin (3 t) : - 5 e^{- 2 t} sin (3 t)

= - 5 e^{- 2 t} sin (3 t)

y = e^{x + 1} + 1

x \frac{d y}{d x} + 2 y = \frac{1}{x ^{3}}

x^{2} + 7 x - 5

\int (\frac{3}{x} - \frac{x x}{4}) d x

z^{^{'}}

What is inverse oflaplace-1{(15)/(s^2+4s+13)} ?
The answer to inverse oflaplace-1{(15)/(s^2+4s+13)} is -5e^{-2t}sin(3t)